# Cofunction Formulas

• Last Updated : 15 May, 2022

A trigonometric cofunction is defined as expressing a trigonometric angle ratio in terms of the other. It illustrates how sine, cosine, tangent, cotangent, secant, and cosecant relate to each other. The cofunction of an angle’s complement is equal to that angle’s trigonometric function. For example, the sine of an angle x is equal to the cosine of the complement of the same angle. Similarly, we can write the cofunction formulas for other ratios as well.

### ASTC rule

The concept of cofunction formulas is based on the trigonometric quadrant table. It follows the ASTC rule, which stands for the “all sin cos tan” rule. It tells which trigonometric ratios are positive in any given quadrant. It further explains that all trigonometric ratios are positive in the first quadrant, only sine and cosecant ratios are positive in the second quadrant, tangent and cotangent are positive in the third quadrant, while cosecant and secant are positive in the fourth quadrant. The first quadrant has angles from 0° to 90°; the second quadrant has 90° to 180°, the third quadrant has 180° to 270°, and the fourth quadrant has 270° to 360°. ### Cofunction Formulas

The cofunction formulas are defined for the angles -x, (90° – x), (90° + x), (180° – x), (180° + x), (270° – x), (270° + x), (360° – x). All trigonometric ratios are evaluated for these angles by substituting the angle x for these. The important point here is that for an angle x added/subtracted by 90° or its multiples, the trigonometric ratio is equal to the positive or negative of its complement ratio. For example, the sine of an angle (90° + x) is cos x. While for an angle x added/subtracted by 180° or its multiples, the trigonometric ratio is equal to positive or negative of its own. For example, the sine of an angle (180° + x) is – sin x. The cofunction formulas are given below in the form of a table:

Derivation

The cofunction formulas can be derived using the sum and difference formulas for various ratios.

Consider the proof of sin (90° – x) = cos x.

We know, sin (A – B) = sin A cos B – cos A sin B.

Here, A = 90° and B = x. So, the formula becomes,

sin (90° – x) = sin 90° cos x – cos 90° sin x

= 1 (cos x) – 0 (sin x)

= cos x – 0

= cos x

Now, consider the proof of cos (90° – x) = sin x.

We know, cos (A – B) = cos A cos B + sin A sin B.

Here, A = 90° and B = x. So, the formula becomes,

cos (90° – x) = cos 90° cos x + sin 90° sin x

= 0 (cos x) + 1 (sin x)

= 0 + sin x

= sin x

This derives the cofunction formulas for sine and cosine ratios. Similarly we can derive the cofunction identities for other ratios as well.

### Sample Problems

Problem 1: Calculate the value of sin 25° cos 75° + sin 75° cos 25°.

Solution:

We know,

sin 25° = cos (90° – 25°) = cos 75°

cos 25° = sin (90° – 25°) = sin 75°

So, the given expression becomes,

sin 25° cos 75° + sin 75° cos 25° = cos 75° cos 75° + sin 75° sin 75°

= cos2 75° + sin2 75°

= 1

Problem 2: Calculate the value of sin 35° cos 65° + sin 65° cos 35°.

Solution:

We know,

sin 35° = cos (90° – 35°) = cos 65°

cos 35° = sin (90° – 35°) = sin 65°

So, the given expression becomes,

sin 35° cos 65° + sin 65° cos 35° = cos 65° cos 65° + sin 65° sin 65°

= cos2 65° + sin2 65°

= 1

Problem 3: Calculate the value of sec 20° cosec 70° – tan 20° cot 70°.

Solution:

cosec 70° = sec (90° – 70°) = sec 20°

cot 70° = tan (90° – 70°) = tan 20°

So, the given expression becomes,

sec 20° cosec 70° – tan 20° cot 70° = sec 20° sec 20° – tan 20° tan 20°

= sec2 20° – tan2 20°

= 1

Problem 4: Calculate the value of cosec 40° sec 50° – cot 40° tan 50°.

Solution:

sec 50° = cosec (90° – 50°) = cosec 40°

tan 50° = cot (90° – 50°) = cot 40°

So, the given expression becomes,

cosec 40° sec 50° – cot 40° tan 50° = cosec 40° cosec 40° – cot 40° cot 40°

= cosec2 40° – cot2 40°

= 1

Problem 5: Calculate the value of tan 1° tan 2° tan 3°… tan 89°.

Solution:

We have,

A = tan 1° tan 2° tan 3°……. tan 89°

= tan 1° tan 2° tan 3°……. tan 87° tan 88° tan 89°

= tan 1° tan 2° tan 3°….. tan 45° ….. tan 87° tan 88° tan 89°

= tan 1° tan 2° tan 3°….. tan 45° ….. cot 3° cot 2° cot 1°

= tan 1° tan 2° tan 3°….. tan 45° ….. (1/tan 3°) (1/tan 2°) (1/tan 1°)

= tan 45°

= 1

Problem 6: Calculate the value of cot 23° cot 41° cot 60° cot 67° cot 49°.

Solution:

cot 23° = tan (90 – 23) = tan 67°

cot 41° = tan (90 – 41) = tan 49°

So, the given expression becomes,

cot 23° cot 41° cot 60° cot 67° cot 49° = tan 67° tan 49° cot 60° cot 67° cot 49°

= tan 67° tan 49° cot 60° (1/tan 67°) (1/tan 49°)

= cot 60°

= 1/√3

Problem 7: Calculate the value of x for the equation, sin 2x = cos (x – 30°).

Solution:

We have,

sin 2x = cos (x – 30°)

Using the identity cos x = sin (90° – x) we get,

sin 2x = sin (90° – (x – 30°))

sin 2x = sin (90° – x + 30°)

sin 2x = sin (120° – x)

=> 2x = 120° – x

=> 3x = 120°

=> x = 40°

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